Problem: You have found the following ages (in years) of all 5 lions at your local zoo: $ 11,\enspace 7,\enspace 3,\enspace 11,\enspace 17$ What is the average age of the lions at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 5 lions at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{11 + 7 + 3 + 11 + 17}{{5}} = {9.8\text{ years old}} $ Find the squared deviations from the mean for each lion. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $11$ years $1.2$ years $1.44$ years $^2$ $7$ years $-2.8$ years $7.84$ years $^2$ $3$ years $-6.8$ years $46.24$ years $^2$ $11$ years $1.2$ years $1.44$ years $^2$ $17$ years $7.2$ years $51.84$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{1.44} + {7.84} + {46.24} + {1.44} + {51.84}} {{5}} $ $ {\sigma^2} = \dfrac{{108.8}}{{5}} = {21.76\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{21.76\text{ years}^2}} = {4.7\text{ years}} $ The average lion at the zoo is 9.8 years old. There is a standard deviation of 4.7 years.